Neglecting all losses, the developed torque (T) of a DC separately excited motor, operating under constant terminal voltage, is related to its output power (P) as under:
Right Answer is:
T ∝ P
SOLUTION
Relation of output power and torque under constant terminal voltage can be given as
The torque developed by a d.c motor is directly proportional to Flux per pole × Armature Resistance i.e
T ∝ ΦIa
T ∝ Ia———1
Output power for the separately excited motor is given as
P = EbIa.
If voltage supply is constant then power is
P ∝ Ia……………. (2)
From equation (1) and (2),
T ∝ P.
or
Consider a pulley of radius r meter acted upon by a circumferential force of F Newton which causes it to rotate at N r.p.m.
The angular speed of the pulley is
ω = 2πN/60 rad/sec
Work done by this force in one revolution
= Force × distance = F × 2πR Joule
The power developed = Work Done/Time
= (F × 2πR)/60/N
= (F × R) × (2πN)/60
The power developed = T × ω watt or P = T ω Watt
From the relation of power developed in the armature is equivalent to mechanical torque developed, is
P = T × ω
Where
P = Output power of separately excited motor and it is given as P = EbIa
ω = Angular speed in rad/sec. and it is given as ω = 2πN ⁄ 60
where N = Speed of motor in rpm,
Eb = induced back emf,
Ia = armature current.
If voltage supply is constant then power is
∴ T∝ power output P will be equal to power developed in the armature.
∴ T ∝ P.